Find the external work done by the system in kcal, when 20 kcal of heat is supplied to the system the increase in the internal energy is 8400J(J=4200J/kcal)
A
22kcal
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B
18kcal
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C
20kcal
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D
19kcal
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Solution
The correct option is A22kcal Weknow,Δd=v+ΔwΔd=−20×4200JΔw=−84000−8400=−92400J